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FLUID MECHANICS – 1FLUID MECHANICS – 1Semester 1 2011 - 2012Semester 1 2011 - 2012
Compiled and modifiedCompiled and modified
byby
Sharma, AdamSharma, Adam
Week – 8, 9 and 10
FLOW IN PIPES- FLOW IN PIPES- CO3CO3
Review
• Conservation of mass (mass balance)
• Continuity Equation
• Bernoulli Equation
• Momentum Equation
3
Objectives• Understand Laminar and Turbulent flow in pipes
• Identify types of flow using Reynolds number
• Explain minor losses and major losses for flow in pipes
• Determine friction factor and major losses using moody chart and simplified Colebrook equation
• Calculate minor losses, major losses, pressure losses and head losses
• Understand equivalent length
• Solve piping system using Bernoulli equation considering all losses.
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1. Introduction• There are two kind of flow
a. Internal flow ( flow in pipes)
b. External flow ( flow over bodies, drag, lift) – will be covered in fluids mechanics 2 (BMM2543)
• Examples of internal flow
1. Water flow in pipes
2. Blood flow
3. Oil and Gas industry
4. Cooling system of a car (Radiator)
5. Air Conditioning (Chilled water system)
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examples & pictures
6
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2. Types of flow• Flow in pipes has three types, Laminar,
Transition, Turbulent
• Laminar flow
―Smooth streamlines (flow)
―Highly ordered motion
―Short in length
― normally appeared in high viscosity flow and small pipe/passage i.e. oil in small pipe
Transition flow
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2. Types of flow (cont’)• Turbulent flow
―Rough streamlines (flow)
―Highly disordered motion
―Most flow in reality is turbulent
―High momentum, thus high friction
• Transition flow
―flow from laminar does not suddenly change to turbulent, it will enter transition flow first
―Normally ignored in calculation
―fluctuation of laminar and turbulent randomly
Is there other method to distinguish these flow?
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• Yes, use Reynolds number, Re
pipe ofdiameter Internal
velocitypipe Average
)(kg/m fluid flowing ofdensity
) m.skg( , viscositydynamic
) sm( , viscositykinematic
avg
3
2
D
V
• Re < 2300, Laminar flow
• 2300 < Re < 4000, Transition flow
• Re > 4000, Turbulent flow
Is there other method to distinguish these flow?(cont)
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• For non-circular pipes, hydraulic diameter, Dh is used for calculating the Reynolds number.
• however, in practical, type of flow depend on smoothness of pipe, vibrations and fluctuation in the flow.
p
AD c
h
4
Perimeter Internal
(internal) Area onalCrosssecti4
Reynolds Number (example)
11
• Q1: Water at 20°C flow with average velocity of 2cm/s inside a circular pipe. Determine flow type if the pipe diameter, a) 2 cm, b) 15 cm, and c) 30 cm
33- 998kg/m m.skg 101.002
C20At 2010) book, (Cengel tableFrom
,
• (c) Turbulent flow59700.001002
0.3)998(0.02)(Re c)
DVave
3980.001002
0.02)998(0.02)(Re a)
DVave • (a) Laminar flow
29880.001002
0.15)998(0.02)(Re b)
DVave • (b) Transition flow
Reynolds Number (example) cont’
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• Q2: Water at 20°C flow in a circular pipe of 3.5 cm diameter. Determine the range for the average velocity so the flow is always transition flow
33- 998kg/m m.skg 101.002
C20At 2010) book, (Cengel tableFrom
,
m/s 0660.035)0998(
0.0010022300.V min,ave
0.001002
.035)0)(998(2300Re min,aveave
imummin
VDV
m/s 1150.035)0998(
0.0010024000.V max,ave
0.001002
.035)0)(998(4000Remaximum
max,aveave VDV
m/s 0.115 m/s 0660 Range aveV.
Reynolds Number (example) cont’
13
• Q3: Air at 35°C flow inside rectangular pipe of 2cmx5cm. Determine the maximum flow velocity for the pipe before the flow enter transition region
sm 106551 C,35At Air
2010), book, (Cengel tableFrom25- .
m/s31310290
1065512300 5
..
.V max,ave
5-101.655
.029)0(2300Re
max,aveave
imummax
VDV
m 02902(0.05)2(0.02)
05)4(0.02)(0.4.
p
AD c
h
4. Entrance Region
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• Between the entrance and fully developed flow
• Uniform velocity profile at entrance
• because of no slip boundary condition, friction at the wall reduce the velocity of flow near the wall
• to conserve mass, velocity at the center increase (compensate velocity decreased near the wall)
4. Entrance Region (cont’)
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• As the fluid move deeper in the pipe, the velocity near the wall decreased further and velocity at center increase (developing velocity profile)
• Both (up & down @ left & right) velocity profile increase till it merge with the other side
• fully developed flow is when the velocity profile stop to develop as it flows deeper inside the pipe
4. Entry Region (cont’)
16
• hydrodynamic entry length, Lh is evaluated from pipe entrance to where wall shear stress achieve 2% of fully developed value or approximately
D.L arminla,h Re050 shorter length for turbulent pipe flow
DL turbulent,h 10
4. Entry Region (example)
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• Q4: Determine the hydrodynamic entry region for Q1 (a & c).
D.L arminla,h Re050
• (c) Turbulent flow59700.001002
0.3)998(0.02)(Re c)
DVave
3980.001002
0.02)998(0.02)(Re a)
DVave • (a) Laminar flow
m 3980020398050Re050 ...D.L arminla,h
m 200201010 ..DL turbulent,h
DL turbulent,h 10
4.1. Turbulence velocity profile VS Laminar velocity profile (fully developed)
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• Velocity profile based on analysis
• Consist of 1 layer
• Small velocity gradient
• The average velocity in fully developed laminar pipe flow is ½ of the maximum velocity
• Velocity profile is based on analysis and empirical
• Consist of 4 layer
• High velocity gradient
avemax Vu 2
5. Losses in piping system
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• There are two type of losses which is major losses and minor losses. Losses are mainly due to friction and obstruction
• Total losses, hL is major losses + minor losses
5.1 Major Losses (pg345)• Major losses, hL, major , is also known pressure
losses or head losses
• The major losses is solely depend on the pipe, nothing else
5.1 Major Losses (cont’)
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• Pressure drop across pipe can be formulated as
• f, Darcy friction factor. There is another friction factor, Fanning friction factor, but will not be covered
• for fully developed laminar flow, friction factor is obtained from combining Darcy equation with Pressure drop.
L
PP
dx
dP 21
2221
328
D
LV
R
LVPPP aveave
2 Rough)or (Smooth loss Pressure
2ave
L
V
D
LfP
g
V
D
Lf
g
Ph aveL
major,L 2 loss, headmajor
2
5.1 Major Losses (cont’)
21
2
32
2
2aveave
L
V
D
Lf
D
LVPP
2
32 aveV
fD
D
Vf ave 32
2
232
DVf ave
64Re f
flow)laminar developedfully for factor (friction Re
64 f
• from the eq. ,friction factor in Laminar flow is independent of roughness
• The head loss represents the additional height that the fluid needs to be raised by a pump in order to overcome the frictional losses in the pipe
5.1 Major Losses (cont’)
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• For fully developed turbulent flow pressure losses or head losses, the equation is more complex
• Unlike laminar flow, friction factor, f , for turbulent flow is depend on internal pipe roughness and Reynolds number base on Colebrook equation.
• Whereas, ε is roughness of pipe and D is pipe diameter.
• However this equation is implicit and cannot be solved directly. Using iteration (numerical method) is possible but the method is tedious
flow) (turbulent Re
512
73 log 2.0
1
f
.
.
D-
f
5.1 Major Losses (cont’)
23
• Hence, there are two option, (a), Moody chart, (b) Modified Colebrook equation
• In Moody Chart, friction factor can be obtain by knowing the roughness ratio, ε/D and Reynolds number
ε/D=0.02
Re=100,000
f =0.025
5.1 Major Losses (cont’)
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• Haaland modified Colebrook Equation (1983).
• Easier to calculate
• The formula has about 2% error compare to original equation
• Formula for major head loss for turbulent is the same
flow)ent for turbul (Modified 73Re
96 log 81
1
111
.
.
D..-
f
d)(simplifie 73Re
96 log 81
-2111
.
.
D..-f
g
V
D
Lfh ave
major,L 2
2
5.1 Major Losses (cont’)
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• In real pipe application (turbulent flow), friction is unwanted because the rougher the surface, the higher the friction
• Old piping system such as Cast Iron, GI, the performance deteriorate through time as corrosion reduce smoothness and size of internal pipe.
• New piping (HDPE), Anti-corrosion metal pipes, smoothness maintain, thus performance is maintained
5.1 Major Losses (example)
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• Q5: Water at 40°C ( = 992.1 kg/m3 and µ = 0.653×10-3 kg/m.s) is flowing steadily in a 5cm-diameter horizontal pipe made of stainless steel at a rate of 300 l/min. Determine the pressure drop, the head loss, and the required pumping power input for flow over a 50m-long section of the pipe.
m/s5524(0.05)
(1000)/60300
2avgavg ./
/
A
QVVAQ
cc
Turbulent ,1937100.000653
.05)0)(992.1(2.55Re
DVavg
0.00004mm50mm0020 ratio, Roughness /.D0.016 101.94 Re & 0.00004 Chart,Moody from 5 f,D
0160 73
000040
193710
96 log81
-2111
..
...f
.
Pa 516092
(2.55)1992
050
500160
2
22
.
..
V
D
LfP ave
L
m 35(9.81)1992
51609.
.g
Ph L
L
Watt25851609601000300 //PQW L
5.1 Major Losses (example cont’)
27
• Q6: Air at 40°C (ρ = 1.127 kg/m3, ν = 1.702×10-6 m2/s) is flowing steadily in a 50cm-diameter horizontal pipe made of plastic at a rate of 30 l/min. Determine the head loss, and the required pumping power input for flow over a 150m-long section of the pipe.
m/s105524(0.5)
(1000)/6030 3
2avgavg .
/
/
A
QVVAQ
cc
Laminar ,74820.00000170
.5)0)(10552(Re
3
.DVavg
08560748
64
Re
64 .f
Pa 1041981)(1.127)(9.10518 56 ..ghP LL
m 10518(2)(9.81)50
)10552(15008560
26
232
..
..
g
V
D
Lfh ave
L
Watt107141041960100030 85 ..//PQW L
28
5.2 Minor Losses• In piping system there are various fittings,
valves, bends, elbows, tees, inlets, exits, enlargements, and contractions in addition to the pipes.
• These component Interrupt the smoothness of flow and cause additional losses
• Usually, these additional losses is called minor losses and smaller than pipe losses .
• Rarely, minor losses will be greater than major losses especially when the component installed frequently along the pipe system (between short distance)
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5.2 Minor Losses (cont’)
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5.2 Minor Losses (cont’)
• usually expressed in terms of the loss coefficient, KL
• KL , is provided by manufacturer and the value varies for different components.
• Consider a component, valve, the pressure losses is losses due to valve minus losses by imaginary pipe section without valves
31
5.2 Minor Losses (cont’)
2
2avg
LL
VKP
• Independent of Re, Reynolds number
• Usually expressed in term of head losses, hL
• Also, In industrial application, the manufacturing data is expressed in terms of equivalent length, Lequiv.
• for this method, simply add Lequiv to the total length of pipe for the total head losses calculation.
• However, the former expression will be used thoroughly
components todue lossminor 2
2
g
VKh avg
LL
32
5.2 Minor Losses (cont’)
Lequivavgequivavg
LL Kf
DL
g
V
D
Lf
g
VKh
22
22
• Component with sharp edge such as sharp edge exit has higher loss coefficient compare to well rounded
• Sharp edge introduce recirculating flow due to fluid unable to make sharp 90° turn especially at high speed
• Same for sudden expansion/contraction
33
5.2 Minor Losses (cont’)
34
5.2 Minor Losses (cont’)
• These loss coefficient depends on the manufacturer data
35
5.2 Minor Losses (cont’)
36
5.2 Minor Losses (cont’)
37
5.2 Minor Losses (cont’)
38
Globe valve
Angle valve
Ball valve
Swing checkvalve
• Total head losses, hL, total
• Calculation of Minor losses is straight forward. If the piping system consist same components such as bends, simply multiply the loss coefficient with the number of the same bends.
g
VK
D
Lfhhh avg
LMinor,LMajor,Ltotal,L 2
2
39
5.2 Minor Losses (cont’)
diameter pipedifferent for different is
diameter) pipe one than more with system pipe(for
2
avg
2avg,
V
g
VK
D
Lfh i
Li
iitotal,L i
• Normally, as an engineer and consultant, piping system for example water storage, sprinkler system, hose reel system is the main concern.
• To solve piping system, extended Bernoulli equation is required which the total losses is placed at the right hand side (at point 2)
• Piping system usually constructed to deliver fluid at higher level or to create a pressurized system
total,Lhzg
V
g
Pz
g
V
g
P 2
222
1
211
2
)(
2
)(
40
6 Piping system
• Two principles in analyzing piping system which are
a) Conservation of mass throughout the system must be satisfied
b) Pressure drop (and thus head loss) between two junctions must be the same for all paths between the two junctions
41
6 Piping system (cont’)
1
2
point 1, Just above water level, P1=?, V1 = ?, Z1 = ?
point 2, Just above water level, P2=?, V2 = ?, Z2 = ?
h
• Q7:A piping system delivering water at 25°C from tank 1 to tank 2. The system consist two 45º, a sharp entrance and a sharp exit. The diameter of the stainless steel pipe is 2cm and length of 55 m. Determine h so that the flowrate is 83.3 L/min.
42
6 Piping system (example)
point 1, Just above water level, P1=0, V1 = 0, Z1 = 0
point 2, Just above water level, P2=0, V2 = 0, Z2 = h
2
1
h=?m45°
45°
Tank 1
Tank 2
43
6 Piping system (example)
OH
22avg
major 2m 2949
(9.81)2
(4.42)
020
550180
2.
..
g
V
D
Lfh ,L
m/s424(0.01)
(60000)383
2avgavg ./.
A
QVVAQ
cc
Turbulent ,98916100.891
.02)0997(4.42)(Re
3-
DVavg
000100.02102 , ratio Roughness
m102mm0020 , Roughness steel, Stainless6
6
.D/
.
0180 73
00010
98916
96 log81
-2111
..
...f
.
44
6 Piping system (example)
total,Lhh 00000
total,Lhzg
V
g
Pz
g
V
g
P 2
222
1
211
2
)(
2
)(
0 0 0 0 Reference point
m 5851.h
OH
22avg
minor 2m 292
(9.81)2
(4.42)140250
2...
g
VKh L,L
m58512922949 ...hhh Minor,LMajor,Ltotal,L
• Q8:A piping system delivering water at 25°C from pressurised tank 1 to tank 2. Initial plastic pipe diameter is 2 cm has a sharp entrance(inlet), fully open globe valve and two 90º smooth flanged bend, and a sudden expansion at the 1/3 of the pipe length. After expansion, Galvanized Iron (GI) pipe diameter is installed with 8 cm diameter and along the GI pipe, there are two 90º miter bend, a fully open angle valve and a sharp exit. The total pipe length is 75 m. Determine gauge pressure at tank 1 required to deliver water at 226 liter/hour.
45
6 Piping system (example)
point 1, Just above water level, P1≠0, V1 = 0, Z1 = 0
point 2, Just above water level, P2=0, V2 = 0, Z2 = h
1
h=25m
2
Tank 2
Tank 1
GI pipe
Plastic pipe
46
6 Piping system (example)m/s20
(0.01)
000)1(3600226
21
avg1avg11 ./
A
QVVAQ
cc
Turbulent ,4475100.891
.02)0997(0.2)(Re
3-
avg11
DV
pipessmooth 0 , Roughness pipe, Plastic ,
factor)friction (turbulent 0390 73
0
4475
96 log81
-2111
1 ..
..f
.
m/s01250(0.04)
(3600000)226
22
avg2avg22 ./
A
QVVAQ
cc
Laminar ,1119100.891
.08)0)(997(0.0125Re
3-
avg22
DV
factor)friction (laminar 0570 1119
64
Re
64 2 .f
47
6 Piping system (example)
g
VKKKK
D
Lfh ,L,L,L,Lplastictotal,L 2
22
avg1expansionsudden bend flangesmooth g.valveinlet
1
11
880080
02011
2
2
22
22
21
expansionsudden ..
.
D
DK ,L
m1240(9.81)2
20880(0.3)21050
020
3750390
2
..
...
/.h plastictotal,L
m000350(9.81)2
0125015(1.1)2
080
375750570
2
I ..
.
/.h Gtotal,L
g
VKKK
D
Lfh ,L,L,LGtotal,L 2
22
avg2exit sharp valveanglebendmiter
2
22I
m1243500003501240I plastic ...hhh Gtotal,Ltotal,Ltotal,L
48
6 Piping system (example)
1243502500001 .g
P
total,Lhzg
V
g
Pz
g
V
g
P 2
222
1
211
2
)(
2
)(
0 0 0Reference point
small very is losses as differenceelevation the
overcome tois 1at tank pressure theofMost
Pa24573081999712435251 ..P
49